彙總表中的行數

我有一個postgres數據庫中的多個表，它們擁有完全獨特的信息。這些信息如果在查詢中正確連接在一起，將會產生我期待的所有可能的組合。我正在尋找的信息是完整的SKU。彙總表中的行數

爲了產生完整的SKU，這個查詢產生所期望的結果： 功能查詢

SELECT 
    materials.code, 
    "part_base_parts".code as part_base_parts_id, 
    shanks.code AS shank_id, 
    measurements.description 
FROM 
    "part_base_parts" 
LEFT JOIN "part_types" ON "part_base_parts"."part_type_id" = "part_types"."id" 
RIGHT JOIN "parts_to_shanks" ON "part_base_parts"."id" = "parts_to_shanks"."part_base_part_id" 
RIGHT JOIN "parts_to_measurements" ON "part_base_parts"."id" = "parts_to_measurements"."part_base_part_id" 
RIGHT JOIN "parts_to_materials" ON "part_base_parts"."id" = "parts_to_materials"."part_base_part_id" 
JOIN materials ON "parts_to_materials"."material_id" = materials."id" 
JOIN shanks ON "parts_to_shanks"."shank_id" = shanks."id" 
JOIN measurements ON "parts_to_measurements"."measurement_id" = measurements."id" 
ORDER BY 
    part_base_parts_id ASC, 
    materials.code ASC, 
    shank_id ASC, 
    measurements.description ASC

鑑於此查詢，我產生32640條記錄（沒有應用索引），用0.82秒查詢時。事情是這樣的......

給定輸出

code part_base_parts_id shank_id description 
AA  5105    A   03.0 
. 
. 32,638 rows in here. 
. 
ST  6939    D   9/16

這是唯一讓我走了一半，雖然。我需要從查詢中取回結果並從每列中生成總計數。所以，我需要有結果將是：

預期的效果

code: AA - ###0 
     ... 
     ST - ###0 
part_base_parts_id: 5105 - ###0 
        ... 
        6939 - ###0 
shank_id: A - ###0 
      ... 
      D - ###0 
description: 03.0 - ###0 
      ... 
      9/16 - ###0

是否有生產「預期效果」從Postgres的的方法嗎？

來源

2016-07-14 Jeff Nichols

如果你想要它們在行中，那麼肯定。

WITH cte AS( 
    SELECT 
     materials.code, 
     "part_base_parts".code as part_base_parts_id, 
     shanks.code AS shank_id, 
     measurements.description 
    FROM 
     "part_base_parts" 
    LEFT JOIN "part_types" ON "part_base_parts"."part_type_id" = "part_types"."id" 
    RIGHT JOIN "parts_to_shanks" ON "part_base_parts"."id" = "parts_to_shanks"."part_base_part_id" 
    RIGHT JOIN "parts_to_measurements" ON "part_base_parts"."id" = "parts_to_measurements"."part_base_part_id" 
    RIGHT JOIN "parts_to_materials" ON "part_base_parts"."id" = "parts_to_materials"."part_base_part_id" 
    JOIN materials ON "parts_to_materials"."material_id" = materials."id" 
    JOIN shanks ON "parts_to_shanks"."shank_id" = shanks."id" 
    JOIN measurements ON "parts_to_measurements"."measurement_id" = measurements."id" 
    ORDER BY 
     part_base_parts_id ASC, 
     materials.code ASC, 
     shank_id ASC, 
     measurements.description ASC  
) 
SELECT key, value, count(*) 
FROM(
    SELECT 'code' AS key, code AS value 
    FROM cte 
    UNION ALL 
    SELECT 'part_base_parts_id', code 
    FROM cte 
    UNION ALL 
    SELECT 'shank_id', shank_id 
    FROM cte 
    UNION ALL 
    SELECT 'description', description 
    FROM cte 
) AS q 
GROUP BY key, value 
ORDER BY key, value

來源

2016-07-14 21:26:20

缺少一個'UNION ALL'。我會添加'ORDER BY key，value'。 – klin

我修復了複製/粘貼錯誤，並按照@klin的建議添加了ORDER BY。 –

該查詢工作@ jakub-kania和@klin！我確實在查詢時間上受到了打擊。從.82秒到最後結果〜3.7秒。我將研究索引並查看要應用它們的表以及這可能如何影響查詢。關於增加表演時間的想法？ –

彙總表中的行數

回答

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